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If $f:\mathbb{C}\rightarrow\mathbb{C}$ is a differentiable function and $f(x+2\pi)=f(x)$ for all $x\in \mathbb{R}$, would $f(z+2\pi)=f(z)$ for all $z\in \mathbb{C}$?

Is there any theorem/lemma concerning this? Are there any examples/counter examples for this?

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If $f$ is holomorphic, then this works. But differentiability is too weak a condition (take for example $f(x+iy) = \sin((1-y)x)$). –  Joel Cohen May 25 '12 at 21:47
    
Do you have a proof? –  Potato May 25 '12 at 21:48
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@Potato : If $f$ is holomorphic, then $g(z) = f(z+2\pi)$ also is. But since $g$ and $f$ are equal on $\mathbb{R}$, the identity theorem (en.wikipedia.org/wiki/Identity_theorem) tells us they are equal on $\mathbb{C}$. –  Joel Cohen May 25 '12 at 21:52
    
You should repost that below as an answer. –  Potato May 25 '12 at 21:56
    
@JoelCohen: This should really be an answer ... –  user26872 May 25 '12 at 21:56
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1 Answer 1

up vote 15 down vote accepted

If $f$ is holomorphic, then $g(z) = f(z+2\pi)$ also is. But since $f$ and $g$ are equal on $\mathbb{R}$, the identity theorem tells us they are equal on $\mathbb{C}$.

Now if you only assume differentiability as a function on $\mathbb{R}^2$, there are counter examples. Take for example $f(x+iy) = \sin(yx)$.

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Arguably, if the OP says that the domain of the function is $\mathbb C$, then differentiable means holomorphic. –  Phira May 25 '12 at 22:13
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@Phira : You're right, this is probably what was implied. I added the last part just to avoid confusion with being differentiable as a function on $\mathbb{R}^2$. –  Joel Cohen May 25 '12 at 22:28
    
Thank you, :) I didnt explicitly mention it, but I would guess that I did actually mean holomorphic. –  Derrick May 25 '12 at 23:46
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